Why is the stalk cohomology of the nearby cycles of a sheaf complex exactly the hypercohomology of the Milnor fiber of that complex?

Question

Let $X$ be a complex manifold and $f\colon X\to\mathbb{C}$ a non-constant holomorphic map. We define $X_t=f^{-1}(t)$. Let $\mathcal{F}^\bullet\in D^b(X)$ and we denote by $\psi_f$ the nearby cycle functor and by $F_x$ the Milnor fiber of $f$ at $x$.

Recall the definition of the nearby cycle functor: Consider the cartesian square (sry for the bad style) \begin{align} & \ \ \tilde{X^*} \longrightarrow \tilde{\mathbb{C}^*}\\ & \tilde{p}\downarrow\ \ \ \ \ \ \ \ \ \downarrow p\\ X_0 \overset{i}{\longrightarrow} & \ \ \ \ X \overset{f}{\longrightarrow} \mathbb{C} \end{align} where $p$ is the universal covering of $\mathbb{C}^*$ embedded in $\mathbb{C}$ and $\tilde{X^*}$ the usual pullback. We define $\psi_f=i^{-1}R\tilde{p}_*\tilde{p}^{-1}$.

I want to show that for all points $x\in X_0$ there is a natural isomorphism $\mathscr{H}^k(\psi_f\mathcal{F}^\bullet)_x\simeq\mathbb{H}^k(F_x,\mathcal{F}^\bullet)$.

I am currently reading

Dimca, Alexandru, Sheaves in topology, Universitext. Berlin: Springer (ISBN 3-540-20665-5/pbk). xvi, 236 p. (2004). ZBL1043.14003.

This result is precisely Proposition 4.2.2. in the book and it says that this can be obtained by direct computation. I tried to do it but I am stuck at one point. Here is what I got until now.

\begin{align} \mathscr{H}^k(\psi_f\mathcal{F}^\bullet)_x & \simeq H^k((\psi_f\mathcal{F}^\bullet)_x)\\ & \simeq H^k((R\tilde{p}_*\tilde{p}^{-1}\mathcal{F}^\bullet)_{i(x)=x})\\ & \simeq H^k(\underset{\underset{x\in U}{\longrightarrow}}{\text{lim}}R\Gamma(U,R\tilde{p}_*\tilde{p}^{-1}\mathcal{F}^\bullet))\\ & \simeq \underset{\underset{x\in U}{\longrightarrow}}{\text{lim}} \mathbb{H}^k(U,R\tilde{p}_*\tilde{p}^{-1}\mathcal{F}^\bullet)\\ & \simeq \underset{\underset{x\in U}{\longrightarrow}}{\text{lim}}\mathbb{H}^k(\tilde{p}^{-1}(U),\tilde{p}^{-1}\mathcal{F}^\bullet) \end{align} Is this even the right way to approach this? Is it okay to pull the limit out of the cohomology?

Answer 1

It seems common for everyone to simply state that the stalk of the nearby cycle is the cohomology of the Milnor fibre and leave it at that. I remember thinking once upon a time that this wasn't obvious, so let me try to write an answer. I will add some extra notation: Let $X_\eta$ denote the complement of $X_0$ in $X$, and $j:X_\eta\to X$ be the inclusion. Write $\pi:\tilde{X_\eta}\to X_\eta$ to be the pullback of the universal cover of $\mathbb{C}^\times$ under $f$. Then $\psi_f=i^\ast j_\ast \pi_\ast \pi^\ast j^\ast$ is the definition of the nearby cycle.

Let's start by mentioning the assumptions on $\mathcal{F}$. We need $\mathcal{F}$ to be constructible. We will also need the geometry of nearby fibres: If $x\in X_0$, then for sufficiently small $\epsilon$, the ball $B_\epsilon (x)$ of radius $\epsilon$ about $x$ has the property that for $\delta$ sufficiently small (compared to epsilon), the map $f$, restricted to $B_\epsilon(x)$, is a fibration over $\{z\in \mathbb{C}\mid 0<|z|<\delta\}$ (Milnor-Le fibration). This will be our nontrivial geometric input. Now choose a stratification for which $\mathcal{F}$ is stratified, and choose $\epsilon$ and $\delta$ such that the above fibration result holds for the restriction of $f$ to every stratum in our stratification. This is possible since the stratification is finite.

Inside this fibration, we obtain an isomorphism between the Milnor fibre $F_x$ and any fibre, and the restriction of $\mathcal{F}$ to any of these fibres "looks the same". (So the cohomology of $F_x$ with coefficients in $\mathcal{F}$ is well-defined).

In the task at hand, we want to compute a stalk of $\psi_f\mathcal{F}$, which means computing the sections of $\psi_f\mathcal{F}$ in a sufficiently small neighbourhood of $x$. The presence of the $i^\ast j_\ast$ factors means that we have to compute the sections of $\pi_\ast\pi^\ast j^\ast \mathcal{F}$ in the intersection of $X_\eta$ with a sufficiently small neighbourhood of $x$.

Now let's turn our attention to the push-pull piece of the construction. Ultimately we will see that this is a contrivance designed to not favour any particular nearby fibre, and its presence is needed in the definition because the base $\mathbb{C}^\times$ is not simply-connected. The relevant fact for our computation is that $\pi$ is a Galois cover, being a pullback of a Galois cover. Let $U$ be a sufficiently small open set in $X_\eta$, so that $\pi^{-1}U\cong \sqcup_{i\in \mathbb{Z}} \tau^i(V)$, where $\pi$ induces a homeomorphism from $V$ to $U$, and $\tau$ is a generator of $\pi_1(\mathbb{C}^\times)$ acting by deck transformations. Then we have $$(\pi_\ast\pi^{\ast}(j^\ast \mathcal{F}))(U)\cong \prod_{i\in \mathbb{Z}} (j^{\ast}\mathcal{F})(U)$$ with each factor coming from one piece in the fibre, and the monodromy group acting by permuting the factors.

We're interested in sections of $\pi_\ast\pi^{\ast}j^\ast \mathcal{F}$ in a small-enough neighbourhood that is a neighbourhood of the central fibre. We assume our neighbourhood is small enough so that $f$ is a fibration in this neighbourhood for all strata relevant to $\mathcal{F}$, which implies $j^\ast \mathcal{F}$ is locally constant. When we take a local section and transport it by a loop around the central fibre, in the interpretation of the section as an element of a product, the component in each factor is shifted (by the indexing $i\mapsto i+1$ in $\mathbb{Z}$). So the component of a local section in a single factor completely determines, and is determined by, a global section.

A local section, over a disk in $\mathbb{C}^\ast$ is precisely the cohomology of the Milnor fibre with coefficients in $\mathcal{F}$, due to $f$ being a fibration here and the disk being contractible. In this way we obtain the desired result, that the stalk of the nearby cycle is the cohomology of the Milnor fibre.